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I found an interesting formula for $\pi$ in David Bailey's 2021 note "A catalogue of mathematical formulas involving $\pi$, with analysis" (PDF link via davidhbailey.com)

Following is the 25-th formula (out of 67, not counting five iterative algorithms) in the paper.

$$\pi=\sum\limits_{n=0}^\infty\dfrac {50n-6}{2^n \binom {3n} {n}}$$

There is something noteworthy about this series :

  1. it is rather simple
  2. The left side is clearly $\pi$ (we don't see $1 \over \pi$, some dirty square root)
  3. The numerator increases very slowly(arithmetic sequence),
    while the denominator increases very quickly.
  4. I also guess that any $\pi$ approximation series whose numerator is an arithmetic sequence is very rare.

Can you tell me where this formula come from?
The paper does not say where this formula came from.

Blue
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imida k
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  • The Credits section of the note mentions that "Some of the summation formulas involving factorials and combinatorial coefficients (i.e.,formulas 19 through 50 [excepting #36 (Borwein & Borwein), #45 & #46 (Guillera), #50 (Almkvist & Guillera)]) were found by Ramanujan; others are due to David and Gregory Chudnovsky." This narrows the search space a bit. – Blue Feb 09 '25 at 16:34
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    Related: https://math.stackexchange.com/q/1331167/323432 (the comments may be useful). – projectilemotion Feb 09 '25 at 17:55
  • @projectilemotion No wonder the formula looked familiar. I'd almost forgotten about that post. – Tito Piezas III Feb 10 '25 at 18:11
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    It seems to belong to an infinite family $$\pi = \sum_{n=0}^\infty \frac{50n-6}{{\binom{3n}{n}2^{n}}}$$ $$3\pi+30 = \sum_{n=0}^\infty \frac{125n^2+7}{{\binom{3n}{n}2^{n}}}$$ $$91\pi+960 = \sum_{n=0}^\infty \frac{3125n^3+129}{{\binom{3n}{n}2^{n}}}$$ $$7263\pi+69780 = \sum_{n=0}^\infty \frac{156250n^4-178}{{\binom{3n}{n}2^{n}}}$$ and so on, but still no proof after 9 years – Tito Piezas III Feb 10 '25 at 18:53

2 Answers2

14

The formula was found in 1974 by Bill Gosper. However, his proof has not been published. Nevertheless, a proof can be found in [1]. I will elaborate on certain aspects of this proof here.

Note that $$\begin{align} \frac{1}{\binom{3n}{n}}&=\frac{(2n)!\cdot n!}{(3n)!}\\&=(3n+1)\cdot \frac{\Gamma(2n+1)\Gamma(n+1)}{\Gamma(3n+2)}\\&=(3n+1)B(2n+1,n+1)\\&=(3n+1)\int_0^1 \left(x^{2}(1-x)\right)^n~dx. \end{align}$$ Also, by differentiating the geometric series $\sum_{n=0}^{\infty} y^n=\frac{1}{1-y}$, one obtains closed forms for $\sum_{n=0}^{\infty} ny^n=\frac{y}{(1-y)^2}$ and $\sum_{n=0}^{\infty} n^2 y^n=\frac{y^2+y}{(1-y)^3}$ (all for $|y|<1$). One gets $$\sum_{n=0}^{\infty} (50n-6)(3n+1)y^n=\frac{2(56y^2+97y-3)}{(1-y)^3}.$$ Using this with $y=\frac{x^2 (1-x)}{2}$, one obtains $$\sum\limits_{n=0}^\infty\dfrac {50n-6}{2^n \binom {3n} {n}}=8\int_0^1 \frac{28x^6-56x^5+28x^4-97x^3+97x^2-6}{(x^3-x^2+2)^3}~dx.$$ The last integral can be done using partial fractions $(x^3-x^2+2)^3=(x+1)^3(x^2-2x+2)^3$: $$\begin{multline}\frac{28x^6-56x^5+28x^4-97x^3+97x^2-6}{(x^3-x^2+2)^3}=\frac{13}{5(x^2-2x+2)}\\+\frac{76x-115}{5(x^2-2x+2)^2}+\frac{156-132x}{5(x^2-2x+2)^3}-\frac{13}{5(x+1)^2}+\frac{12}{5(x+1)^3}. \end{multline}$$ Alternatively, it can be done using Ostrogradsky's method (particularly elegant here!): $$\begin{multline} \int \frac{28x^6-56x^5+28x^4-97x^3+97x^2-6}{(x^3-x^2+2)^3}~dx=\frac{x^5-29x^4+37x^3-x^2-8x}{2(x^3-x^2+2)^2}\\+\frac{1}{2}\int \frac{x+1}{x^3-x^2+2}~dx, \end{multline}$$ which gives $$\frac{x^5-29x^4+37x^3-x^2-8x}{2(x^3-x^2+2)^2}+\frac{\arctan(x-1)}{2}+C.$$

[1] Almkvist, G., Krattenthaler, C., & Petersson, J. (2003). Some new formulas for $\pi$. Experimental mathematics, 12(4), 441-456. Link: https://www.arxiv.org/pdf/math.NT/0110238.

projectilemotion
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  • Any chance of proving the rest of the family? It's been siting idle in MSE for 9 long years. I've added it as a comment above, and I'll ask it as a separate question if you have ideas how to prove the rest. – Tito Piezas III Feb 10 '25 at 19:00
  • I think that the proofs are essentially the same. For instance, for the second identity in your comment you would get that $$\sum_{n=0}^\infty \frac{125n^2+7}{{\binom{3n}{n}2^{n}}}=\int_0^1 \sum_{n=0}^{\infty} (125n^2+7)(3n+1)y^n=\int_0^1 \frac{264y^3+1479y^2+500y+7}{(1-y)^4}~dx$$ with $y=\frac{x^2(1-x)}{2}$ (WolframAlpha gives the correct value $3\pi+30$ for this integral). The same seems to be possible for your identities with order $7,11$. For example, by a similar computation using the beta function $$\frac{1}{\binom{7n}{2n}}=(7n+1)\int_0^1 (x^5 (1-x)^2)^n~dx.$$ – projectilemotion Feb 11 '25 at 00:20
  • In general, $$\sum_{n=1}^\infty \dfrac{P(n)}{{\displaystyle \binom{mn}{2n}2^{n-1}}}=2\int_0^1 \sum_{n=1}^{\infty} P(n)(mn+1)\left(\frac{x^{m-2} (1-x)^{2}}{2}\right)^{n}~dx,$$ if I'm not mistaken (a closed form for the sum for a given polynomial $P(n)$ is easy to find). – projectilemotion Feb 11 '25 at 00:25
  • Can you convert your comment for $m=3$ into an answer for the other post? But I think it may be slightly more complicated since for $m=7$ there seems to be infinitely many polynomials $P(n)$ with deg $6$. And second, the closed-form for $m\neq 4n+3$ doesn't seem to be of form $a\pi+b$. I'll ask it as a new question. – Tito Piezas III Feb 11 '25 at 05:09
  • Are you sure? I don't think I am answering your questions in your post. Both of your questions are more or less about finding these identities, whereas all that I have done in my comments is I provided an algorithm to prove them (for any $m\geq 3$) by computing the sum. In all cases computing the sum reduces to computing the integral of a rational function (it becomes more difficult the higher $m$ is). – projectilemotion Feb 11 '25 at 13:22
  • I'm not sure how to give a complete proof of this. However here's a heuristic about why this could be true (cf. [1]). In all cases, you end up with an integral of the form $$\int_0^1 \frac{Q(x)}{(x^2(1-x)^{m-2}-2)^{d+2}}~dx$$ where $d=\deg(P)$. A nice way to get a $\pi$ is for the anti-derivative to have an $\arctan(x)$ or $\arctan(x-1)$. For this you want $f_m(x)=x^2(1-x)^{m-2}-2$ to have either $x^2+1$ or $(x-1)^2+1$ as a factor, i.e. $f_m$ should have a zero at $i$ or $1+i$. This is the case for all $m=4v+3$, indeed $$f_{4v+3}(1+i)=0.$$ This is not the case for $m=4v+1$! – projectilemotion Feb 11 '25 at 14:12
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    I think that will do. A partial answer that is on the right track may eventually lead yourself or others to the full answer. I’ve seen it happen. I will msg you when I edit and trim my post. – Tito Piezas III Feb 11 '25 at 17:23
  • I've eliminated the second question. The first one should be more manageable. Kindly check the post. – Tito Piezas III Feb 13 '25 at 02:52
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For the fun of it, I played with $$\sum_{n=0}^\infty \frac{a n+b}{\binom{3 n}{n}}x^n=\sum_{n=0}^\infty (a n+b)\,\frac{\Gamma (n+1)\, \Gamma (2 n+1)}{\Gamma (3 n+1)}x^n$$

Using generalized hypergeometric functions $$S_1=\sum_{n=0}^\infty n\,\frac{\Gamma (n+1)\, \Gamma (2 n+1)}{\Gamma (3 n+1)}x^n=\frac{x}{3} \, \, _3F_2\left(\frac{3}{2},2,2;\frac{4}{3},\frac{5} {3};\frac{4 x}{27}\right)$$ $$S_2=\sum_{n=0}^\infty \frac{\Gamma (n+1)\, \Gamma (2 n+1)}{\Gamma (3 n+1)}x^n=\, _3F_2\left(\frac{1}{2},1,1;\frac{1}{3},\frac{2} {3};\frac{4 x}{27}\right)$$

If $x=\frac 12$ $$S_1=\frac{1}{6} \, _3F_2\left(\frac{3}{2},2,2;\frac{4}{3},\frac{5} {3};\frac{2}{27}\right)=0.2050269345227246880860924\cdots$$ $$S_2=\,_3F_2\left(\frac{1}{2},1,1;\frac{1}{3},\frac{2} {3};\frac{2}{27}\right)=1.184959012091073527640329\cdots$$

Searching for integer $a$ and $b$ (a double loop), the only combination to obtain $\pi$ gives the result

Claude Leibovici
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